A255855 Least k > 0 such that gcd(k^n+5, (k+1)^n+5) > 1, or 0 if there is no such k.

1, 0, 1, 5, 1, 533360, 1, 55, 1, 7, 1, 796479131355665831357, 1, 41, 1, 5, 1, 3775, 1, 42296, 1, 7, 1, 653246700175064613889, 1, 21, 1, 5, 1, 1619, 1, 42842, 1, 7, 1, 2945, 1, 323371, 1, 5, 1, 1102221, 1, 633524110177, 1, 7, 1

Offset: 0

Author

M. F. Hasler, Mar 08 2015

Keywords

• nonn
• hard

Comments

See A118119, which is the main entry for this class of sequences.

Examples

			For n=1, gcd(k^n+5, (k+1)^n+5) = gcd(k+5, k+6) = 1, therefore a(1)=0.
For n=2k, see formula.
For n=3, we have gcd(5^3+5, 6^3+5) = 13, and the pair (k,k+1)=(5,6) is the smallest which yields a GCD > 1, therefore a(3)=5.
		

Crossrefs

Cf. A118119, A255832.

Cf. A255852, A255853, A255854, A255856, A255857, A255858, A255869.

Programs

Mathematica

A255855[n_] := Module[{m = 1}, While[GCD[m^n + 5, (m + 1)^n + 5] <= 1, m++]; m]; Join[{1, 0}, Table[A255855[n], {n, 2, 10}]]

PARI

a(n,c=5,L=10^7,S=1)->for(a=S,L,gcd(a^n+c,(a+1)^n+c)>1&&return(a))

Python

from sympy import primefactors, resultant, nthroot_mod
from sympy.abc import x
def A255855(n):
    if n == 0: return 1
    k = 0
    for p in primefactors(resultant(x**n+5,(x+1)**n+5)):
        for d in (a for a in sorted(nthroot_mod(-5,n,p,all_roots=True)) if pow(a+1,n,p)==-5%p):
            k = min(d,k) if k else d
            break
    return int(k) # Chai Wah Wu, May 08 2024

Formula

a(2k)=1 for k>=0, because gcd(1^(2k)+5,2^(2k)+5) = gcd(6,4^k-1) = 3.

Extensions

a(11)-a(46) from Hiroaki Yamanouchi, Mar 12 2015